Introduction to Probability and Statistics for Engineers and Scientists

106 Chapter 4:Random Variables and Expectation

EXAMPLE 4.3g Suppose thatp(x,y), the joint probability mass function ofXandY,is
given by

p(0, 0)=.4, p(0, 1)=.2, p(1, 0)=.1, p(1, 1)=.3.

Calculate the conditional probability mass function ofXgiven thatY=1.

SOLUTION We first note that

P{Y= 1 }=

∑

x

p(x,1)=p(0, 1)+p(1, 1)=.5

Hence,

P{X= 0 |Y= 1 }=

p(0, 1)

P{Y= 1 }

=2/5

P{X= 1 |Y= 1 }=

p(1, 1)

P{Y= 1 }

=3/5 ■

IfX andY have a joint probability density functionf(x,y), then the conditional
probability density function ofX, given thatY =y, is defined for all values ofysuch
thatfY(y)>0, by

fX|Y(x|y)=

f(x,y)

fY(y)

To motivate this definition, multiply the left-hand side bydxand the right-hand side by
(dx dy)/dyto obtain

fX|Y(x|y)dx=

f(x,y)dx dy

fY(y)dy

≈

P{x≤X≤x+dx,y≤Y≤y+dy}

P{y≤Y≤y+dy}

=P{x≤X≤x+dy|y≤Y≤y+dy}

In other words, for small values ofdxanddy,fX|Y(x|y)dxrepresents the conditional
probability thatXis betweenxandx+dx, given thatYis betweenyandy+dy.
The use of conditional densities allows us to define conditional probabilities of events
associated with one random variable when we are given the value of a second random
variable. That is, ifXandYare jointly continuous, then, for any setA,

P{X∈A|Y=y}=

∫

A

fX|Y(x|y)dx